Question

Differancial equation of y=axcube + bxsquare

Answer 1

Answer:

ANSWER

y=Ax

3

+Bx

2

dx

dy

=3Ax

2

+2Bx

x

1

dx

dy

=3Ax+2B -- Equation (1)

Differentiating again, we get

x

2

−1

dx

dy

+

x

1

dx

2

d

2

y

=3A --Equation (2)

Putting values of A&B in Equation 1 & 2 respectively in y=Ax

3

+Bx

2

, we get

As A=

3

1

(

x

1

dx

2

d

2

y

−

x

2

1

dx

dy

)

2B=

x

1

dx

dy

−3Ax

2B=

x

1

dx

dy

−

dx

2

d

2

y

+

x

1

dx

dy

=(

x

2

dx

dy

−

dx

2

d

2

y

)

B=(

x

1

dx

dy

−

2

1

dx

2

d

2

y

)

Now, Ax

3

+Bx

2

=

3

1

x

2

dx

2

d

2

y

−

3

1

x

dx

dy

+x

dx

dy

−

2

1

x

2

dx

2

d

2

y

Ax

3

+Bx

2

=y=

6

−1

x

2

dx

2

d

2

y

+

3

2

x

dx

dy

6y=x(4

dx

dy

)−x

2

dx

2

d

2

y

or

x

2

dx

2

d

2

y

−4x

dx

dy

+6y=0

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